The function given is:

$f(x) = \frac{5}{2x^2 - 6x - 8}$

Step 1: Find the Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (since the function will be undefined there). To find these points, we set the denominator equal to zero and solve for $x$:

$2x^2 - 6x - 8 = 0$

Divide the equation by 2 to simplify:

$x^2 - 3x - 4 = 0$

Now, factor the quadratic:

$(x - 4)(x + 1) = 0$

So, the solutions are:

$x = 4 \quad \text{and} \quad x = -1$

These are the vertical asymptotes: $x = 4$ and $x = -1$.

Step 2: Find the Horizontal Asymptote

Horizontal asymptotes are determined by the degrees of the numerator and the denominator. The degree of the numerator (5) is 0, and the degree of the denominator $2x^2 - 6x - 8$ is 2.

• If the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is $y = 0$.

Thus, the horizontal asymptote is $y = 0$.

Conclusion

• Vertical asymptotes: $x = -1$ and $x = 4$
• Horizontal asymptote: $y = 0$

Would you like me to plot the asymptotes on the graph provided?

Here are 5 related questions you might be interested in:

1. How do slant (oblique) asymptotes differ from horizontal asymptotes?
2. How can you find asymptotes for more complex rational functions?
3. What happens to the graph near vertical asymptotes?
4. Can a rational function have more than one horizontal asymptote?
5. How does end behavior relate to horizontal asymptotes?

Tip: To factor quadratics easily, look for factors of the constant term that add up to the middle term's coefficient.